Let s be a convergent series in our valuation metric space. (By convergent I mean cauchy convergent.) If sn floats above ε for large n, then the partial sum through n, minus the partial sum through n-1, exceeds ε for large n, and the sequence of partial sums cannot be cauchy convergent. In other words, the terms of a convergent series must approach 0. This holds for every metric space, but the converse does not.

Remember the harmonic series? Its terms approach 0, but it does not converge. Well - in a valuation metric, s converges iff its terms approach 0. The valuation of the difference between two partial sums is at least the smallest valuation of all the terms between those two partial sums. This valuation increases as the terms approach 0. Therefore the partial sums approach each other, and s is convergent.

Scale a convergent sequence by the constant c in the field F. This adds the valuation of c to the valuation of each term. The valuation of the difference between any two terms si and sj is increased by the valuation of c. As we move out along the sequence, the valuations of the differences still approach infinity, and terms cluster together. Therefore c*s is also convergent.

If s and t are convergent sequences, then the valuation of (si+ti) - (sj+tj) is at least the valuation of si-sj or the valuation of ti-tj. Both these valuations approach infinity, hence s+t is a convergent sequence.

Verify that a linear combination of convergent series is convergent. Apply the above paragraphs to the sequence of partial sums.

Let q have a positive valuation and build the geometric series sn = qn. The terms approach 0, so s is convergent. Scale s by 1-q and find another convergent series. The partial sums are 1-qn+1, hence the scaled series converges to 1. This means s approaches 1/(1-q), just like any other geometric series.

Build a valuation ring based on the prime 7, and things can get a bit strange. Set q = 7 and our geometric series becomes 1+7+49+343+…, And this approaches -1/6. It's hard to believe, but add 1/6 to the first few partial sums and watch the valuations increase.

Can we multiply two sequences together, term by term, to get something convergent?

Assume, beyond n, si-sj has a small metric. Let a = si and let a+u = sj, where u has a high valuation. Similarly, let b = ti and let b+v = tj. Now consider (a+u)*(b+v)-ab. Its valuation is at least that of av, bu, or uv. This increases along with u and v, hence the product sequence is convergent.

The inverse of a sequence is also convergent. Represent two nearby terms by a and a+u and consider 1/(a+u) - 1/a. This has the valuation of u/a2, which increases with u. Combine with the previous paragraph, and the term by term quotient of two convergent sequences is convergent, provided the divisor sequence does not approach 0.

The "points" in the completion of our metric space are cauchy sequences. To be precise, a point is an equivalence class of cauchy sequences that converge to the same thing. If s and t are cauchy, and s-t converges to 0, then s and t represent the same point in the completion of our metric space. This is a bit like 3.699999… and 3.700000… representing the same real number.

The original elements of F embed in the completion in the usual way. If x is in F then the sequence x,x,x,x,x… is in the completion of F.

Now there is a lot of stuff you have to do if you want to be rigorous. You need to show, for instance, that addition is well defined in the completion. We already know s+t gives a convergent sequence, which is a point in the completion. But what if s is replaced with some other sequence that also converges to s? Does s+t still land in the same equivalence class?

Add a sequence e to s and consider (s+e)+t. Here e approaches zero, so s and s+e represent the same point in the completion. By associativity, (s+e)+t = e+(s+t). The new result is really the same as s+t, adjusted by a sequence that approaches 0. Thus addition is well defined.

If c is a scalar and s a sequence, and e is a sequence that approaches 0, c*(s+e) = c*s+c*e. Since c*e drops to 0, multiplication by c is well defined. Set c = -1 and subtraction is well defined.

Step gently into multiplication by considering a sequence s and another sequence e that approaches 0. Move out beyond some large n, and let a = si and a+u = sj. Of course u has a high valuation, and the terms ei and ej also have a high valuation - as high as you like. Now consider (a+v)ej - aei. Rewrite this as a(ej-ei) + vej. Since e is cauchy the first term drops to zero. Since e and v both have high valuation, the second term also approaches 0. Any sequence in the completion, multiplied by a sequence that approaches 0, produces another sequence that approaches 0. Symbolically, s*0 = 0.

Now consider the product of two arbitrary sequences s*t. Add a sequence e to s, and the product sequence is shifted by e*t. If e approaches 0 then e*t approaches 0. All the sequences that represent s give the same result, namely s*t, and multiplication is well defined.

You can also prove that addition and multiplication are commutative and associative, and the latter distributes over the former, and 0 and 1 are the respective identities. In other words, the completion forms a ring. Furthermore, the original space embeds in the completion as a subring, and as a subspace with the induced topology. (I'll leave all this to you.)

Finally, assume the sequence s does not approach 0, and consider the term by term inverse of s. We already know this converges, so add the sequence e to s. The difference between 1/(s+e) and 1/s is e/(s*(s+e)). If s is convergent, yet does not approach 0, then it is, beyond some n, bounded away from 0. (This is a lemma that you may want to verify.) So the valuation of the terms of s is, eventually, bounded below by some value b. Go out further, so that the valuation of the terms of e is always beyond b. Now the denominator is no worse then b2 in valuation, and the numerator keeps increasing. The metric approaches 0, and the sequence 1/(s+e) has the same limit as the sequence 1/s. Division is well defined, and the completion forms a field.

I ran through this pretty fast, but it's all very similar to theorems you've seen before, when we completed the rationals to produce the reals.

The rationals are all p-adic, because every metric space embeds in its completion.

Is the field of p-adic numbers algebraically closed? Turn the page and find out.